Let $\mathcal{P} = A_1 A_2 \cdots A_{n}$. All indices in this solution will be cyclic – i.e., $A_{n+k}$ is considered to be the same as $A_k$. Consider the circumcircle of the $n$-gon. It has a radius of $R = \frac{1}{2 \sin \frac{\pi}{n}}$. The area of triangle $A_i A_j A_k$ is then \[\frac{A_i A_j \cdot A_j A_k \cdot A_k A_i}{4R}\] by a well known formula for area of a triangle. Since the length of $A_i A_j$ only depends on $j – i$, we can rewrite this area as \[\frac{A_n A_{j-i} \cdot A_n A_{k-j} \cdot A_n A_{i-k+n}}{4R}.\]

Thus, the resulting product is \[\frac{\prod_{r=1}^{n-1} (A_n A_{r})^{s_r}}{(4R)^{\binom{n}{3}}},\] where $s_r$ is the number of $j-i$, $k-j$, or $i-k+n$ ($1 \leq i < j < k \leq n$) which are equal to $r$.

The number $s_r$ can also be computed: it is equal to $n(n-r-1)$ by some simple constructive counting. Thus, the needed product is \[\frac{\prod_{r=1}^{n-1} (A_n A_{r})^{n(n-r-1)}}{(4R)^{\binom{n}{3}}}.\] Let $P = \prod_{r=1}^{n-1} (A_n A_{r})^{n(n-r-1)}$. Then, by a index change $r \mapsto n – r$ in the product and a use of the fact that $A_n A_r = A_{n} A_{n-r}$, we find also that $P = \prod_{r=1}^{n-1} (A_n A_{r})^{n(r-1)}$. Multiplying these two expressions together, we find \[P^2 = \prod_{r=1}^{n-1} (A_n A_{r})^{n(n-2)} =\left( \prod_{r=1}^{n-1} A_n A_{r}\right)^{n(n-2)}.\] Because for a regular polygon inscribed in a unit circle, the product of the distances from one vertex to all the other vertices is the number of sides of the polygon (cf. this result), we have $\prod_{r=1}^{n-1} A_n A_{r} = nR^{n-1}$, so $P^2 = (n R^{n-1})^{n(n-2)}$, so $P = \sqrt{n}^{n(n-2)} R^{3 \binom{n}{3}}$.

The final answer is then \[\frac{\prod_{r=1}^{n-1} (A_n A_{r})^{n(n-r-1)}}{(4R)^{\binom{n}{3}}} = \frac{P}{(4R)^{\binom{n}{3}}}.\] Plugging in the known formulas for $R$ and $P$, this can be simplified to \[\boxed{\frac{\sqrt{n}^{n(n-2)}}{(4\sin\frac{\pi}{n})^{2 \binom{n}{3}}}.}\]